# Mod 6 Multiple choice (1 Viewer)

#### NexusRich

##### Member
Please provide working out on this question and explain how neutralisation works with polyprotic acids, thanks in advance

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#### Vaibhav123456

##### New Member
Just by looking at the question you know that LiOH is in excess. Thus, using n = cv, n(excess LiOH) = n(LiOH total) - n(H2SO4) = 0.0035 mol.

Since LiOH --> Li+ + OH-, n(OH) = 0.0035 mol.

To find [OH] = 0.0035/0.08 = 0.04375 mol/L.

pOH = -log(0.04375) = 1.359.

pH = 14 - pOH = 12.64.

Neutralisation works the same with polyprotic acids, it just changes the mole ratios im pretty sure. However, for the concentrations, if a weak acid was in excess im pretty sure they would give more information for equlibrium calculations.

#### CM_Tutor

##### Moderator
Moderator
The equation for the reaction is

H2SO4 + 2 LiOH ---> Li2SO4 + 2 H2O

We have n(H2SO4) = cV = 0.0500 x 50.00 x 10-3 = 2.50 x 10-3 mol

And n(LiOH) = cV = 0.200 x 30.00 10-3 = 6.00 x 10-3 mol

And n(LiOH) used to neutralise H2SO4 = (2 / 1) x 2.50 x 10-3 = 5.00 x 10-3 mol

So, n(LiOH) after reaction = n(LiOH) initially - n(LiOH) reacted = 6.00 x 10-3 - 5.00 x 10-3 = 1.00 x 10-3 mol

So, after reaction, [OH-] = [LiOH] = n / V = 1.00 x 10-3 / [(50.00 + 30.00) x 10-3] = 0.0125 mol L-1 (3 sig fig)

So, assuming the temperature is 25 degC, pH = 14 + log_10 [OH-] = 12.097 (3 dec. pl.)